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Svetlanka [38]
3 years ago
12

Which of the following equations is non-linear?

Mathematics
1 answer:
lakkis [162]3 years ago
5 0

2x + 6y = 14y - 19x^2 + 12 is a non-linear equation

Step-by-step explanation:

Lets define a linear equation first.

A linear equation is an equation in which there is no variable with exponent greater than 1 or the degree of the equation is 1.

So,

<u>x + 12 = -8x + 10 - 2y</u>

The equation is a linear equation because the degree of the equation is 1.

<u>x = 8x + 19 - 10y</u>

The equation is a linear equation because the degree of the equation is 1.

<u>2x + 6y = 14y - 19x^2 + 12</u>

The equation involve a term with exponent 2 which makes the degree of the equation 2 making it a quadratic equation

<u>2x + 13y + 14x - 7 = 16y - 3</u>

The equation is a linear equation because the degree of the equation is 1.

Hence,

2x + 6y = 14y - 19x^2 + 12 is a non-linear equation

Keywords: Linear, quadratic

Learn more about equations at:

  • brainly.com/question/5102020
  • brainly.com/question/5147732

#LearnwithBrainly

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3 years ago
The cost of 300 metres optic fibre is $200 what will be the cost if 1 metre of optic fibre.
Elden [556K]

The answer is:  " $0.67 / metre " ;  
                        or, write as:  " [ \frac{2}{3} dollar] per metre."
____________________

<u>Step-by-step explanation</u>:

This is a unit rate problem:
   Find the cost in dollars per [single unit—in this case: "metre(s)", "<em>m</em>" ;
_________________
 \frac{200 dollars}{300m} = \frac{? dollars}{1m} ;  Solve for the "?" <u>amount of dollars</u>.
_________________
Method 1)  "Cross-factor multiply" :
<u>Note</u>:  <u>Given</u><u>:</u> " \frac{a}{b} = \frac{c}{d} ;  [b\neq 0;  d\neq 0] " ;

 then:   ⟷  " ad = bc " ;
_________________

So:  " \frac{200 dollars}{300m} = \frac{? dollars}{1m} " ;

  then:   ⟷  " (200*1) = (300* ?) ;

     → "  (200) = (300  * ?)" ;

     ↔ " (300 * ?) = 200 " ;  

Let "x" refer to the "?" ; the "<u>unknown value</u>" (<u>in dollars</u>);

                for which we are trying to solve.

⇒ " 300x = 200 " ;  Solve for "x" ;
     → Divide each side of the equation by: "300" ;

       to isolate "x" on the "left-hand side" of the equation; & to              solve  for "x" ;
   ⇒  300x/ 300 = 200/300 ;

    To get:  " x = \frac{2}{3} dollars" ;  

= $0.666666666... ;

round to: $0.67 .

{since: " 1 dollar = $1.00 = 100 cents" ;
and since: "\frac{2}{3} " ;

= ["2 ÷ 3" = 0.666666...." ] ;  

⇒  round to:  "0.67 " ;  

and write as:
__________

⇒  $0.67 / metre ;   or, write as:
    " [ \frac{2}{3} dollar] per metre."
_________

Method 2)  <u>Note</u>:
\frac{200 dollars}{300m} = \frac{? dollars}{1m} ;

⇔   = "  \frac{200}{300}  \frac{dollars}{metre} " ;
      {<u>Note</u>:  "[200÷300]" can be simplified by "canceling out" the two (2) zeros in both the numerator and the denominator.}.

   = " \frac{2 dollars}{3m} ";
_________

  =  i.e. " \frac{\frac{2}{3} dollars}{metre} " or;  \frac{0.67 dollars} {m} ;  or:  <u>$0.67 / metre</u> .

         ⇒  {since:  " \frac{2}{3} of 1 dollar" = " \frac{2}{3} dollars of 100 cents" ;
        ⇒  {since:  "1 dollar = 100 cents"} ;

           ⇒ {and: " \frac{2}{3} " = 0.6666666666666... " } ;  

 → round to:  "0.67 " ; and thus: $0.67 / metre.
_________
Note that using either of these 2 (two) methods result in the same answer.
               Hope this is helpful!  Best wishes to you within your academic pursuits!
             _________________

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Step-by-step explanation:

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